$\overline{AC}$ is $24$ units long $\overline{BC}$ is $10$ units long $\overline{AB}$ is $26$ units long What is $\cot(\angle BAC)?$ $A$ $C$ $B$ $24$ $10$ $26$
Answer: $\cot(\angle BAC) = \dfrac{1}{\tan(\angle BAC)}$ How can we find $\tan(\angle BAC)$ SOH CAH TOA angent = pposite over djacent Opposite $= \overline{BC} = 10$ Adjacent $= \overline{AC} = 24$ $\tan(\angle BAC) = \dfrac{10}{24}$ $\cot(\angle BAC) = \dfrac{1}{\tan(\angle BAC)} = \dfrac{24}{10}$